Among all the students at a certain high school, the probability of pi

Question

Among all the students at a certain high school, the probability of picking a left-handed student is  \frac{1}{4} , and the probability of picking a student who is learning Spanish is  \frac{2}{3}  Which of the following could be the probability of picking a student who is either left-handed or learning Spanish or both?

I.  \frac{2}{3} 
II.  \frac{3}{4} 
III.  \frac{5}{6}

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

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nick1816 · Accepted Answer

Assume total students = LCM(3,4)=12

N(l)= \frac{1}{4}*12=3 
 N(s)= \frac{2}{3}*12=8

We need the minimum and maximum value of

= N(l) + N(s) -

1. For the maximum value, minimize  .
Minimum possible value of  =0

= N(l) + N(s)-0
 = 3+8=11

2. For the minimum value, maximize  .
Maximum possible value of  =3

= N(l) + N(s)-3
 = 3+8-3=8

probability of picking a student who is either left-handed or learning Spanish or both = P

\frac{8}{12} ≤ P ≤ \frac{11}{12}

All 3 options lie in the range.

chetan2u · Answer

Use 2 by 2 matrix
Let the total number be 120 as the fractions have denominator 3 and 4
............S........NS
L...........b....... a..... 120*\frac{1}{4}=30 
R..........c.........d....  90 
Total.....80......40.... 120

Now we are looking at 
 \frac{(a+b+c)}{120} 
Least value will be when a=0 and b=30 and c=50.... \frac{80}{120}=\frac{2}{3} 
Greatest value will be when b=0 and a=30 and c=80... \frac{(80+30)}{120}=\frac{11}{12}

All the fractions are within this range, so E

Let us test

I.  \frac{2}{3} ....b=30, a=0 and c=50
II.  \frac{3}{4} .... b=20, a=10 and c=60
III.  \frac{5}{6} ...b=10, a=20 and c=70

riagarg07 · Answer

Among all the students at a certain high school, the probability of picking a left-handed student is  \frac{1}{4} , and the probability of picking a student who is learning Spanish is  \frac{2}{3}  Which of the following could be the probability of picking a student who is either left-handed or learning Spanish or both?

I.  \frac{2}{3} 
II.  \frac{3}{4} 
III.  \frac{5}{6}

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

Hi Bunuel,

Could you please share the solution?

sanjitscorps18 · Answer

Considering the fractions let's assume the total students to be 12.
hence Left handers would be 1/4 of 12 that is 3
hence Spanish learners would be 2/3 of 12 that is 8

Now, creating the 2 x 2 matrix. In this matrix we have considered the number of people who neither learn Spanish nor are left-handed as 'a'.

---------------------Left-------|--------Not Left-----|-----Total
Spanish------------a - 1------|-------- 9 - a -------|-------8
Not Spanish-------4 - a ------| -------- a ---------|-------4
Total--------------- 3 ---------|----------9 ---------|-------12

Basis the matrix we determine that we need the value of (a-1) + (9-a) + (4 -a) = 12 - a
Hence, we need the probability of (12 -a) / 12, hence, we need the value of 'a'

Now we can see from the matrix that the maximum value of 'a' can be 4.

By that virtue, the probability of picking a student who is either left-handed or learning Spanish or both is greater than or equal to (12 - 4)/12 = 2/3

Required P(e) >=2/3. Checking the options, we get all the values are possible.

IMO Option E

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Among all the students at a certain high school, the probability of pi

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Among all the students at a certain high school, the probability of pi

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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